Bound these Sines

First, assume $\beta$ and $\gamma$ are constants. The expression turns into $c_1\sin \alpha +c_2$ for some constants $c_1$ and $c_2$ which is a linear expression in $\sin\alpha$ . This means that the extremes are at the endpoints of the range of $\sin\alpha$ , which is $-1,1$ . The same reasoning can be applied to $\beta$ and $\gamma$ to obtain $\sin\alpha=-1\text{ or }1$ $\sin\beta=-1\text{ or }1$ $\sin\gamma=-1\text{ or }1$

Thus we have a few cases to check: $\begin{aligned}(\sin\alpha,\sin\beta,\sin\gamma)=(1,1,1)\qquad&\text{Case 1}\\ \text{or }(1,1,-1)\qquad&\text{Case 2}\\ \text{or }(1,-1,-1)\qquad&\text{Case 3}\\ \text{or }(-1,-1,-1)\qquad&\text{Case 4} \end{aligned}$

Plugging these values in, we see that $(1,1,1)\implies 3$ $(1,1,-1)\implies 3$ $(1,-1,-1)\implies -5$ $(-1,-1,-1)\implies 3$

Thus, $M=3$ and $m=-5$ . Therefore the answer is $\boxed{8}$ .

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The equality cases for maximum is: $(\alpha,\beta,\gamma)=\left(\dfrac{\pi}{2}+2\pi n_{\alpha}, \dfrac{\pi}{2}+2\pi n_{\beta}, \dfrac{\pi}{2}+2\pi n_{\gamma}\right)$ $(\alpha,\beta,\gamma)=\left(\dfrac{\pi}{2}+2\pi n_{\alpha}, \dfrac{\pi}{2}+2\pi n_{\beta}, \dfrac{3\pi}{2}+2\pi n_{\gamma}\right)\quad\text{and permutations}$ $(\alpha,\beta,\gamma)=\left(\dfrac{3\pi}{2}+2\pi n_{\alpha}, \dfrac{3\pi}{2}+2\pi n_{\beta}, \dfrac{3\pi}{2}+2\pi n_{\gamma}\right)$

The equality case for minimum is: $(\alpha,\beta,\gamma)=\left(\dfrac{\pi}{2}+2\pi n_{\alpha}, \dfrac{3\pi}{2}+2\pi n_{\beta}, \dfrac{3\pi}{2}+2\pi n_{\gamma}\right)\quad\text{and permutations}$

where $n_{\alpha},n_{\beta},n_{\gamma}$ are integers.

For simplicity, let $a = sin(\alpha), b = sin(\beta), c = sin(\gamma)$ . Then $-1 \le a,b,c \le 1$ .

Let $f(a,b,c)$ be the expression we need to minimize. Then, $f(a,b,c) = (a+b+c) + (ab+ac+bc) - 3abc$ We can rewrite it as $f(a,b,c) = (1+a)(1+b)(1+c) - 1 - 4abc$ .

Note that each of $1+a,1+b,1+c$ are $\ge 0$ . Therefore, $(1+a)(1+b)(1+c) \ge 0$ By similar logic, $- 4abc \ge -4$ Therefore, $f(a,b,c) \ge 0 - 1 - 4 = -5$ This value is achieved because $f(1,-1,-1) = -5$ (i.e. when $\alpha = \frac{\pi}{2}, \beta = \gamma = \frac{-\pi}{2}$ ). Thus $m = -5$ .

Now let's consider $3 - f(a,b,c)$ . We can rewrite it as follows: $3 - f(a,b,c) = 3 - (ab+ac+bc) + 3abc - (a+b+c)$ $= (1-ab) + (1-ac) + (1-bc) + (abc-a)+(abc-b)+(abc-c)$ $= \left[(1-bc) - a(1-bc) \right] + \left[(1-ac) - b(1-ac) \right] + \left[(1-ab) - c(1-ab) \right]$ $= (1-bc)(1-a) + (1-ac)(1-b) + (1-ab)(1-c)$

Note that each of $1-a,1-b,1-c,1-bc, 1-ac, 1-ab$ are $\ge 0$ . Therefore this whole expression is $\ge 0$ .

Thus $3 - f(a,b,c) \ge 0$ , so $f(a,b,c) \le 3$ . This value is achieved since $f(1,1,1) = 3$ (i.e. when $\alpha = \beta = \gamma = \frac{\pi}{2}$ . Hence $M = 3$ .

Therefore $M - m = \boxed{8}$ .